Question

如何转换看起来像这样的词典列表：

[{'id':2, 'risk':'a'}, 
 {'id':1, 'risk':'a'}, 
 {'id':32,'risk':'aa'},
 {'id':2, 'risk':'aa'}, 
 {'id':7, 'risk':'a'}, 
 {'id':7, 'risk':'b'}]

进入一个元组字典，在排序后将如下所示：

{1:('a', ), 2:('a','aa'), 7:('a','b'), 32:('aa', )}

Answer 1

您可以使用defaultdict自动创建元组，然后迭代list_of_dicts：

list_of_dicts = [{'id':2, 'risk':'a'}, 
 {'id':1, 'risk':'a'}, 
 {'id':32,'risk':'aa'},
 {'id':2, 'risk':'aa'}, 
 {'id':7, 'risk':'a'}, 
 {'id':7, 'risk':'b'}]

from collections import defaultdict

dict_of_tuples = defaultdict(tuple)

for dct in list_of_dicts:
    dict_of_tuples[dct['id']] += (dct['risk'],)

结果是：

>>> dict_of_tuples
defaultdict(<type 'tuple'>, {32: ('aa',), 1: ('a',), 2: ('a', 'aa'), 7: ('a', 'b')})

如果您想要一个排序字典：

>>> from collections import OrderedDict
>>> OrderedDict(sorted(dict_of_tuples.items()))
OrderedDict([(1, ('a',)), (2, ('a', 'aa')), (7, ('a', 'b')), (32, ('aa',))])

Answer 2

dict.setdefault方法简化了这类问题：

>>> lod = [{'id':2, 'risk':'a'}, 
           {'id':1, 'risk':'a'}, 
           {'id':32,'risk':'aa'},
           {'id':2, 'risk':'aa'}, 
           {'id':7, 'risk':'a'}, 
           {'id':7, 'risk':'b'}]
>>> dot = {}
>>> for d in lod:
        idnum, risk = d['id'], d['risk']
        dot.setdefault(idnum, []).append(risk)

>>> dot
{32: ['aa'], 1: ['a'], 2: ['a', 'aa'], 7: ['a', 'b']}

您也可以使用 collections.defaultdict 创建相同的效果，但这不会创建常规字典，并且需要了解工厂函数和零参数构造函数。

Answer 3

通常人们在类似情况下忘记使用dict.get方法。所以使用基本的python函数：

list_of_dicts = [{'id':2, 'risk':'a'}, 
    {'id':1, 'risk':'a'}, 
    {'id':32,'risk':'aa'},
    {'id':2, 'risk':'aa'}, 
    {'id':7, 'risk':'a'}, 
    {'id':7, 'risk':'b'}]

final_dict = {}

for item in list_of_dicts:
    final_dict[item['id']] = final_dict.get(item['id'], tuple()) + (item['risk'],)


>> {1: ('a',), 2: ('a', 'aa'), 7: ('a', 'b'), 32: ('aa',)}

Answer 4

以更长的方式做到这一点。不像agf的解决方案那么整洁，但它可以正常工作

new_dict = {}
for x in my_dict_list: 
    m = new_dict.get(x['id'],())
    m += (x['risk'],)
    new_dict[x['id']] = m

输入

my_dict_list = [{'id':2, 'risk':'a'}, 
                {'id':1, 'risk':'a'}, 
                {'id':32,'risk':'aa'},
                {'id':2, 'risk':'aa'}, 
                {'id':7, 'risk':'a'}, 
                {'id':7, 'risk':'b'}]

输出

>>> new_dict
{32: ('aa',), 1: ('a',), 2: ('a', 'aa'), 7: ('a', 'b')}

将词典列表转换为元组词典

4 个答案: